Transform N to Minimum possible value

• Difficulty Level : Expert
• Last Updated : 07 Jul, 2021

Given two numbers and N and D. Apply any of two below operations to N

2. change N to digitsum(N), where digitsum(N) is the sum of digits of N

The task is to transform N to the minimum possible value. Print the minimum possible value of N and the number of times the given operations applied(any one of them). The number of operations must be minimum.
Examples:

Input : N = 2, D = 1
Output : 1 9
Perform Type1 operation 8 times and Type2 operation 1 time
Input : N = 9, D = 3
Output : 3, 2
Apply one type1 operation first and then type2 operation

Prerequisites:
1. Digital Root (repeated digital sum) of the given large integer
2. Numbers in a Range with given Digital Root
Approach :
Let Dr(x) be a function defined for integer x as :

• Dr(x) = x, if 0 <= x <= 9
• else, Dr(x) = Dr(Sum-of-digits(x))

The function Dr(x) is the digital root of a number x.

• Dr(a+b) = Dr(Dr(a) + Dr(b))
• Dr(ab) = Dr(Dr(a) * Dr(b))

Important observation : The minimum value is always the minimum over : Dr(N + kD) for some non-negative integer k.

Dr(N + kD) = Dr(Dr(N) + Dr(kD))          (1)

Now, Dr(kd) = Dr(Dr(k) * Dr(D))
Possible values of Dr(k) are 0, 1, 2…9, given by numbers k=0, 1, 2…9

Dr(x) = Dr(Sum-of-digits(x))             (2)

• The minimum value for N is equal to the minimum value for Sum-of-digits(N). If we reduce this answer once and add D, the minimum value that can be obtained wouldn’t change. So, if it is required to perform a reduce operation and then an add operation, then we can do the add operation and then the reduce operation without affecting the possible roots we can reach. This is evident from combination of formulae (1) and (2)

• So, we can do all add operations first, all reduce operations later, and reach any number that can be possibly reached by any set of operations. Using the above claims, we can prove the minimum possible value is the minimum of Dr(N + kD) where 0 <= k <= 9.

• To find the minimum number of steps, note that the relative order of the add and Sum-of-digits operations does affect the answer. Also, note that the Sum-of-digits function is an decreases extremely fast.

• Any number <= 1010 goes to a number <= 90, any number <= 90 goes to something <= 18 and so on. In short, any number can be reduced to its digital root in <= 5 steps.

• Via this, we can prove that the value of the minimum steps can never be greater than 15. This is a loose upper bound, not the exact one.

• Use brute force recursion algorithm, that at each step branches in 2 different directions, one x = Sum-of-digits(x), the other being x = x+D, but only until a recursion depth of 15. In this way, we stop after exploring 215 different ways.

Below is the implementation of the above approach:

C++

 // CPP program to transform N to the minimum value #include using namespace std;   // Initialising the answer int min_val = INT_MAX; int min_steps = 0;   // Function to find the digitsum int sumOfDigits(int n) {     string s = to_string(n);       int sum = 0;       // Iterate over all digits and add them     for (int i = 0; i < s.length(); i++) {         sum += (s[i] - '0');     }           // Return the digit su,     return sum; }   // Function to transform N to the minimum value void Transform(int n, int d, int steps) {     // If the final value is lesser than least value     if (n < min_val) {         min_val = n;         min_steps = steps;     }       // If final value is equal to least value then check     // for lesser number of steps to reach this value     else if (n == min_val) {         min_steps = min(min_steps, steps);     }       // The value will be obtained in less than 15 steps as     // proved so applying normal recursive operations     if (steps < 15) {         Transform(sumOfDigits(n), d, steps + 1);         Transform(n + d, d, steps + 1);     } }   // Driver code int main() {     int N = 9, D = 3;           // Function call     Transform(N, D, 0);           // Print the answers     cout << min_val << " " << min_steps;           return 0; }

Java

 // JAVA program to transform N to the minimum value import java.util.*;   class GFG{    // Initialising the answer static int min_val = Integer.MAX_VALUE; static int min_steps = 0;    // Function to find the digitsum static int sumOfDigits(int n) {     String s = String.valueOf(n);        int sum = 0;        // Iterate over all digits and add them     for (int i = 0; i < s.length(); i++) {         sum += (s.charAt(i) - '0');     }            // Return the digit su,     return sum; }    // Function to transform N to the minimum value static void Transform(int n, int d, int steps) {     // If the final value is lesser than least value     if (n < min_val) {         min_val = n;         min_steps = steps;     }        // If final value is equal to least value then check     // for lesser number of steps to reach this value     else if (n == min_val) {         min_steps = Math.min(min_steps, steps);     }        // The value will be obtained in less than 15 steps as     // proved so applying normal recursive operations     if (steps < 15) {         Transform(sumOfDigits(n), d, steps + 1);         Transform(n + d, d, steps + 1);     } }    // Driver code public static void main(String[] args) {     int N = 9, D = 3;            // Function call     Transform(N, D, 0);            // Print the answers     System.out.print(min_val+ " " +  min_steps);        } }   // This code is contributed by 29AjayKumar

Python3

 # Python3 program to transform N to the minimum value import sys;   # Initialising the answer min_val = sys.maxsize; min_steps = 0;   # Function to find the digitsum def sumOfDigits(n) :       s = str(n);       sum = 0;       # Iterate over all digits and add them     for i in range(len(s)) :         sum += (ord(s[i]) - ord('0'));           # Return the digit su,     return sum;   # Function to transform N to the minimum value def Transform(n, d, steps) :     global min_val;global min_steps;           # If the final value is lesser than least value     if (n < min_val) :         min_val = n;         min_steps = steps;       # If final value is equal to least value then check     # for lesser number of steps to reach this value     elif (n == min_val) :         min_steps = min(min_steps, steps);           # The value will be obtained in less than 15 steps as     # proved so applying normal recursive operations     if (steps < 15) :         Transform(sumOfDigits(n), d, steps + 1);         Transform(n + d, d, steps + 1);   # Driver code if __name__ == "__main__" :       N = 9; D = 3;           # Function call     Transform(N, D, 0);           # Print the answers     print(min_val, min_steps);       # This code is contributed by Yash_R

C#

 // C# program to transform N to the minimum value using System;   class GFG{    // Initialising the answer static int min_val = int.MaxValue; static int min_steps = 0;    // Function to find the digitsum static int sumOfDigits(int n) {     string s = n.ToString();        int sum = 0;        // Iterate over all digits and add them     for (int i = 0; i < s.Length; i++) {         sum += (s[i] - '0');     }            // Return the digit su,     return sum; }    // Function to transform N to the minimum value static void Transform(int n, int d, int steps) {     // If the final value is lesser than least value     if (n < min_val) {         min_val = n;         min_steps = steps;     }        // If final value is equal to least value then check     // for lesser number of steps to reach this value     else if (n == min_val) {         min_steps = Math.Min(min_steps, steps);     }        // The value will be obtained in less than 15 steps as     // proved so applying normal recursive operations     if (steps < 15) {         Transform(sumOfDigits(n), d, steps + 1);         Transform(n + d, d, steps + 1);     } }    // Driver code public static void Main(string[] args) {     int N = 9, D = 3;            // Function call     Transform(N, D, 0);            // Print the answers     Console.Write(min_val+ " " +  min_steps); } }   // This code is contributed by Yash_R

Javascript



Output:

3 2

Time Complexity : My Personal Notes arrow_drop_up
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